函数y2的图象上． （Ⅰ）若??13，??12，求函数y2的解析式；
（Ⅱ）在（Ⅰ）的条件下，若函数y1与y2的图象的两个交点为A，B，当△ABM的面积为
112时，求t的值；
（Ⅲ）若0?????1，当0?t?1时，试确定T，?，?三者之间的大小关系，并说明理由．
2009参考答案及评分标准
评分说明：
1．各题均按参考答案及评分标准评分．
2．若考生的非选择题答案与参考答案不完全相同但言之有理，可酌情评分，但不得超过该题所分配的分数．
一、选择题：本大题共10小题，每小题3分，共30分.
1．A 2．B 3．B 4．C 5．D 6．A 7．A 8．B 9．D 10．C 二、填空题：本大题共8小题，每小题3分，共24分.
11．2 12．2
13．正方形（对角线互相垂直的四边形均可）
?1? 14．?0，D 2 3 C 15．56，80，156.8
E 1 16．60；13
B A 17．21 2 3
18．①3，4（提示：答案不惟一）；
②裁剪线及拼接方法如图所示：图中的点E可以是以BC为直径的半圆上的任意一点（点B，C除外）.BE，CE的长分别为两个小正方形的边长. 三、解答题：本大题共8小题，共66分 19．本小题满分6分
?5x?1?2x?5，①解：??
x?4?3x?1②?由①得x?2， ························································································································ 2分 由②得，x??52 ···················································································································· 4分
?原不等式组的解集为x?2 ································································································ 6分
20．本小题满分8分.
解：（Ⅰ）这个反比例函数图象的另一支在第三象限. ························································· 1分 因为这个反比例函数的图象分布在第一、第三象限， 所以m?5?0，解得m?5. ································································································ 3分
y （Ⅱ）如图，由第一象限内的点A在正比例函数y?2x的图象上，
y=2x 2x0??x0?0?，则点B的坐标为?x0，0?， 设点A的坐标为?x0，?S△OAB?4，?12x0·2x0?4，解得x0?2（负值舍去）.
A O B x 4?.·?点A的坐标为?2，·········································································································· 6分
又?点A在反比例函数y??4?m?52m?5x的图象上，
，即m?5?8.
8x?反比例函数的解析式为y?. ··························································································· 8分
21．本小题满分8分.
解（Ⅰ）法一：根据题意，可以画出如下的树形图：
3 2 1 第一个球
1 2 第二个球 2 1 3 3
从树形图可以看出，摸出两球出现的所有可能结果共有6种； 法二：根据题意，可以列出下表：
第二个球 （1，3） （2，3）
3 （1，2） （3，2）
2
（2，1） （3，1） 1
1 2 3 第一个球
从上表中可以看出，摸出两球出现的所有可能结果共有6种. ············································· 4分 （Ⅱ）设两个球号码之和等于5为事件A.
3?，2?. 摸出的两个球号码之和等于5的结果有2种，它们是：?2，?3，?P?A??26?13. ··················································································································· 8分
P
22．本小题满分8分.
解（Ⅰ）?PA是⊙O的切线，AB为⊙O的直径， C ?PA⊥AB.
A B ??BAP?90°.
O
??BAC?30°，
??CAP?90°??BAC?60°． ················································································· 2分 又?PA、PC切⊙O于点A，C. ?PA?PC.
?△PAC为等边三角形. ??P?60°. ··························································································································· 5分
（Ⅱ）如图，连接BC， 则?ACB?90°．
在Rt△ACB中，AB?2，?BAC?30°，
?AC?AB·cos?BAC?2cos30°=?△PAC为等边三角形， ?PA?AC.
3.
?PA?3. ···························································································································· 8分
23．本小题满分8分
解：如图，过C点作CD垂直于AB交BA的延长线于点D. ············································· 1分 在Rt△CDA中，AC?30，?CAD?180°??CAB?180??120??60?. ···················· 2分
?CD?AC·sin?CAD?30·sin60°?153. AD?AC·cos?CAD?30·cos60°=15.
C 又在Rt△CDB中，
?BC?70，BD2?BC-CD，
22D A B ?BD?70?1532??2?65. ··························································································· 7分
?AB?BD?AD?65?15?50，
答：A，B两个凉亭之间的距离为50m. ················································································ 8分 24．本小题满分8分.
30?4x，24x?260x?600； ·解（Ⅰ）20?6x，································································· 3分
2（Ⅱ）根据题意，得24x?260x?600??1??22?1?············································ 5分 ??20?30. ·
3?整理，得6x?65x?50?0. 解方程，得x1?则2x?53，3x?5652，x2?10（不合题意，舍去）.
．
53答：每个横、竖彩条的宽度分别为cm，
52cm. ································································· 8分
y B C O A 图①
225．本小题满分10分.
解（Ⅰ）如图①，折叠后点B与点A重合， 则△ACD≌△BCD.
设点C的坐标为?0，m??m?0?. 则BC?OB?OC?4?m. 于是AC?BC?4?m.
22y B D x C O B′ 图②
x D B C y D x O B′ 图③
在Rt△AOC中，由勾股定理，得AC?OC?OA， 即?4?m??m?2，解得m?22232.
?3??点C的坐标为?0，?. ········································································································· 4分
?2?
（Ⅱ）如图②，折叠后点B落在OA边上的点为B?, 则△B?CD≌△BCD. 由题设OB??x，OC?y， 则B?C?BC?OB?OC?4?y，
在Rt△B?OC中，由勾股定理，得B?C?OC?OB?.
??4?y??y?x，
222222即y??18x?2····················································································································· 6分
2由点B?在边OA上，有0≤x≤2，
? 解析式y??18x?22?0≤x≤2?为所求.
? ?当0≤x≤2时，y随x的增大而减小， ?y的取值范围为
32≤y≤2. ······················································································· 7分
（Ⅲ）如图③，折叠后点B落在OA边上的点为B??，且B??D∥OB. 则?OCB????CB??D.
??OCB????CBD，有CB??∥BA. 又??CBD??CB??D，?Rt△COB??∽Rt△BOA. 有
OB??OA?OCOB，得OC?2OB??. ···················································································· 9分
在Rt△B??OC中，
设OB???x0?x?0?，则OC?2x0. 由（Ⅱ）的结论，得2x0??18x20?2，
?x0?0，?x0??8?45. 解得x0??8?45．85?16. ·?点C的坐标为0，···················································································· 10分
??26．本小题满分10分.
解（Ⅰ）?y1?x，y2?x?bx?c，y1?y2?0，
?x??b?1?x?c?0. ··································································································· 1分
22将??213，??12分别代入x??b?1?x?c?0，得
2211?1??1??b?1??c?0，?b?1??c?0， ????????3322????
解得b?16，c?16.
2?函数y2的解析式为y2?x?2621256x?16． ····································································· 3分
（Ⅱ）由已知，得AB?，设△ABM的高为h，
?S△ABM?12AB·h?h?1123，即2h?1144.
根据题意，t?T?16162h，
56161144由T?t?56562t?，得?t?114412t??.
当t?2t?1616??时，解得t1?t2?5?125122；
5?122当t?2t??144时，解得t3?，t4?.
?t的值为
5?25?2，，. ······················································································ 6分 1212125（Ⅲ）由已知，得
????b??c，????T????t??T????t??222?b??c，T?t?bt?c.
2??t???b?，
??t???b?，
2???????b??c?????b??c?，化简得??????????b?1??0.
?0?????1，得????0， ?????b?1?0.
有??b?1???0，??b?1???0. 又0?t?1，?t???b?0，t???b?0，
?当0?t≤a时，T≤?≤?；
当??t≤?时，??T≤?；
当??t?1时，????T. ································································································· 10分